Exercicio de metodos dos elementos finitos

Exercícios
Determinar a função de forma para os elementos
1)
1 2 3 4
2)
1 2 3 4 5 a) Pela condição
Nixi=1Nixj=1
b) Usando o polinômio de Lagrange
Resolução:
1.a) ξ1=-1 ξ2=-13 ξ3=13 ξ4=1
Ni=ai+bi∙ξ+ci∙ξ2+di∙ξ3
Para o nó 1:
N1ξ1=a1+b1∙ξ1+c1∙ξ12+d1∙ξ13=1
N1ξ2=a1+b1∙ξ2+c1∙ξ22+d1∙ξ23=0
N1ξ3=a1+b1∙ξ3+c1∙ξ32+d1∙ξ33=0
N1ξ4=a1+b1∙ξ4+c1∙ξ42+d1∙ξ43=0 Para o nó 2:
N2ξ1=a2+b2∙ξ1+c2∙ξ12+d2∙ξ13=0
N2ξ2=a2+b1∙ξ2+c2∙ξ22+d2∙ξ23=1
N2ξ3=a2+b2∙ξ3+c2∙ξ32+d2∙ξ33=0
N2ξ4=a2+b2∙ξ4+c2∙ξ42+d2∙ξ43=0 Para o nó 3:
N3ξ1=a3+b3∙ξ1+c3∙ξ12+d3∙ξ13=0
N3ξ2=a3+b3∙ξ2+c3∙ξ22+d3∙ξ23=0
N3ξ3=a3+b3∙ξ3+c3∙ξ32+d3∙ξ33=1
N3ξ4=a3+b3∙ξ4+c3∙ξ42+d3∙ξ43=0 Para o nó 4:
N4ξ1=a4+b4∙ξ1+c4∙ξ12+d4∙ξ13=0
N4ξ2=a4+b4∙ξ2+c4∙ξ22+d4∙ξ23=0
N4ξ3=a4+b4∙ξ3+c4∙ξ32+d4∙ξ33=0
N4ξ4=a4+b4∙ξ4+c4∙ξ42+d4∙ξ43=1 N1=-116+116∙ξ-916∙ξ2+916∙ξ3=116∙-1+ξ+9∙ξ2-9∙ξ3
N2=916-2716∙ξ-916∙ξ2+2716∙ξ3=916∙1-3∙ξ-ξ2+3∙ξ3
N3=916+2716∙ξ-916∙ξ2-2716∙ξ3=916∙1+3∙ξ-ξ2-3∙ξ3
N4=-116-116∙ξ+916∙ξ2+916∙ξ3=116∙-1-ξ+9∙ξ2+9∙ξ3

1.b)
Niξe=j=i, j≠1Nξ-ξjξi-ξj
N1ξ1=ξ-ξ2ξ1-ξ2∙ξ-ξ3ξ1-ξ3∙ξ-ξ4ξ1-ξ4
N1ξ1=ξ+13-1+13∙ξ-13-1-13∙ξ-1-1-1=116∙-1+ξ+9∙ξ2-9∙ξ3
N2ξ2=ξ-ξ1ξ2-ξ1∙ξ-ξ3ξ2-ξ3∙ξ-ξ4ξ2-ξ4
N2ξ2=ξ+1-13+1∙ξ-13-13-13∙ξ-1-13-1=916∙1-3∙ξ-ξ2+3∙ξ3
N3ξ3=ξ-ξ1ξ3-ξ1∙ξ-ξ2ξ3-ξ2∙ξ-ξ4ξ3-ξ4
N3ξ3=ξ+113+1∙ξ+1313+13∙ξ-113-1=916∙1+3∙ξ-ξ2-3∙ξ3
N4ξ4=ξ-ξ1ξ4-ξ1∙ξ-ξ2ξ4-ξ2∙ξ-ξ3ξ4-ξ3
N4ξ4=ξ+11+1∙ξ+131+13∙ξ-131-13=116∙-1-ξ+9∙ξ2+9∙ξ3

2.a) ξ1=-1 ξ2=-12 ξ3=0 ξ4=12 ξ5=1
Ni=ai+bi∙ξ+ci∙ξ2+di∙ξ3+ei∙ξ4
Para o nó 1:
N1ξ1=a1+b1∙ξ1+c1∙ξ12+d1∙ξ13+e1∙ξ14=1
N1ξ2=a1+b1∙ξ2+c1∙ξ22+d1∙ξ23+e1∙ξ24=0
N1ξ3=a1+b1∙ξ3+c1∙ξ32+d1∙ξ33+e1∙ξ34=0
N1ξ4=a1+b1∙ξ4+c1∙ξ42+d1∙ξ43+e1∙ξ44=0
N1ξ5=a1+b1∙ξ5+c1∙ξ52+d1∙ξ53+e1∙ξ54=0
1 -1 1 -1 1 1 -1214-18116 1 0 0 0 0 1 12 14 18 116 1 1 1 1